# Proof of theorem...

• January 26th 2010, 05:13 PM
Aileys.
Proof of theorem...
It's the conditional Bayes theorem:

Some conditions, blah blah

Then $\bar{E}[X|\mathcal{G}] = \frac{E[\Lambda X|\mathcal{G}]}{E[\Lambda|\mathcal{G}]}$

I'm fine with the bulk of the proof, it's just that it starts off by defining

$Y=\frac{E[\Lambda X|\mathcal{G}]}{E[\Lambda|\mathcal{G}]}$ if $E[\Lambda|\mathcal{G}]>0$ and Y = 0 otherwise

Then it says we need to show $Y = \bar{E}[X|\mathcal{G}]$

But I don't really see that $Y=\frac{E[\Lambda X|\mathcal{G}]}{E[\Lambda|\mathcal{G}]}$

In fact if $E[\Lambda|\mathcal{G}] = 0$ Then the RHS of the formula is undefined! So.... what's going on here? :S

Thanks for any help!
• January 31st 2010, 02:40 PM
Aileys.
up from page 3...